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choli [55]
3 years ago
12

Factored completely, the expression 3x^2 - 3x - 18 is equivalent to

Mathematics
1 answer:
Dima020 [189]3 years ago
4 0

Answer:

3(x−3)(x+2)

Step-by-step explanation:

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Harry saved $100 each week for 8 weeks. He earned $48 on his savings of $800. What interest did Harry earn for every $100?
ch4aika [34]
<h2>Explanation:</h2><h2></h2>

In this problem, we know that Harry saved $100 each week for 8 weeks. In other words, he saved a total amount of money:

\text{Amount of money saved}=8\times 100=\$800

We know that he earned $48 on his savings of $800, so for every $100 the interest (I) he earns is:

I=\frac{48}{8}=\$6

So, in conclusion Harry did earn $6 in interest for every $100

8 0
4 years ago
What is the slope that passes through (-4,4) and (-6,6)
Bad White [126]

Answer:(2,2)

Step-by-step explanation:

7 0
4 years ago
Given: x + 2y = -6.<br><br><br><br> Solve for y.
zavuch27 [327]

Answer:

The answer is option 3.

Step-by-step explanation:

The steps are :

x + 2y =  - 6

2y =  -x - 6

y =  \frac{ - x - 6}{2}

7 0
3 years ago
Read 2 more answers
Joe is the proud owner of Joe's Old Time BBQ Pit. He sells two types of BBQ sauce - Hot N' Spicy and Mild Mannered. Each gallon
oksano4ka [1.4K]

Answer:

P=$254

Step-by-step explanation:

let Hot N' Spicy be x

and Mild Mannered be y

  the objective function is

38x+35y=P

the constraints are

1. tomato sauce

2x+3y=18--------1

2. hot peppers

2x+y=10--------2

solving 1 and 2 we have

using substitution method

2x+y=10--------2

y=10-2x

put y=10-2x in eqn 1

2x+3(10-2x)=18

2x+30-6x=18

2x-6x=18-30

-4x=-12

x=12/4

x=3

put x=3 in eqn 2 we have

2(3)+y=10

6+y=10

y=10-6

y=4

He should make 3 gallons of Hot N' Spicy

and 4 gallons Mild Mannered

38(3)+35(4)=P

114+140=P

254=P

P=$254

5 0
3 years ago
Can anyone figure this out?
Verizon [17]

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ N(\stackrel{x_1}{-3}~,~\stackrel{y_1}{10})\qquad A(\stackrel{x_2}{6}~,~\stackrel{y_2}{3})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ NA=\sqrt{(6+3)^2+(3-10)^2}\implies NA=\sqrt{130} \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_2}{6}~,~\stackrel{y_2}{3})\qquad D(\stackrel{x_1}{6}~,~\stackrel{y_1}{-1}) \\\\\\ AD=\sqrt{(6-6)^2+(-1-3)^2}\implies AD=4 \\\\[-0.35em] ~\dotfill


\bf D(\stackrel{x_1}{6}~,~\stackrel{y_1}{-1})\qquad N(\stackrel{x_1}{-3}~,~\stackrel{y_1}{10}) \\\\\\ DN=\sqrt{(-3-6)^2+(10+1)^2}\implies DN=\sqrt{202}


now that we know how long each one is, let's plug those in Heron's Area formula.


\bf \qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a=\sqrt{130}\\ b=4\\ c=\sqrt{202}\\[1em] s=\frac{\sqrt{130}+4+\sqrt{202}}{2}\\[1em] s\approx 14.81 \end{cases} \\\\\\ A=\sqrt{14.81(14.81-\sqrt{130})(14.81-4)(14.81-\sqrt{202})} \\\\\\ A=\sqrt{324}\implies A=18

5 0
4 years ago
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